LeetCode 19 — Remove Nth Node From End of List（C++ Java Python）

题目：http://oj.leetcode.com/problems/remove-nth-node-from-end-of-list/

Given a linked list, remove the nth node from the end of list and return its head.

For example,

Given linked list: 1->2->3->4->5, and n = 2.

After removing the second node from the end, the linked list becomes 1->2->3->5.

Note:

Given n will always be valid.

Try to do this in one pass.

题目翻译：

给定一个链表，删除倒数第n个节点，并返回头结点。

例如，

   给定链表：1->2->3->4->5，n = 2。

   删除倒数第2个节点之后，该链表变为1->2->3->5。

注意：

给定的n始终是有效的。

尝试一遍完成。
分析：
        采用快慢指针的思想，间隔n，当快指针到达尾部时，慢指针的下一个结点即为要删除的结点。注意考虑边界情况。

C++实现：

/**
* Definition for singly-linked list.
* struct ListNode {
*     int val;
*     ListNode *next;
*     ListNode(int x) : val(x), next(NULL) {}
* };
*/
class Solution {
public:
ListNode *removeNthFromEnd(ListNode *head, int n) {
// Note: The Solution object is instantiated only once and is reused by each test case.
ListNode *fast = head;
ListNode *slow = head;

for(int i = 0; i < n; ++i)
{
fast = fast->next;
}

if(fast != NULL)
{
while(fast->next != NULL)
{
fast = fast->next;
slow = slow->next;
}

ListNode *tmp = slow->next;
slow->next = tmp->next;
delete tmp;

return head;
}
else
{
ListNode *tmp = head;
head = head->next;
delete tmp;

return head;
}
}
};

Java实现：

/**
* Definition for singly-linked list.
* public class ListNode {
*     int val;
*     ListNode next;
*     ListNode(int x) {
*         val = x;
*         next = null;
*     }
* }
*/
public class Solution {
public ListNode removeNthFromEnd(ListNode head, int n) {
// Note: The Solution object is instantiated only once and is reused by each test case.
ListNode fast = head;
ListNode slow = head;

for (int i = 0; i < n; ++i) {
fast = fast.next;
}

if (fast != null) {
while (fast.next != null) {
fast = fast.next;
slow = slow.next;
}
slow.next = slow.next.next;

return head;
} else {
return head.next;
}
}
}

Python实现：

# Definition for singly-linked list.
# class ListNode:
#     def __init__(self, x):
#         self.val = x
#         self.next = None

class Solution:
# @return a ListNode
def removeNthFromEnd(self, head, n):
fast = head
slow = head

for i in range(n):
fast = fast.next

if fast != None:
while fast.next != None:
fast = fast.next
slow = slow.next

slow.next = slow.next.next

return head

else:
return head.next

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